# Example 10 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 10 Find the term independent of x in the expansion of 32𝑥2 − 13𝑥6,x > 0. Step1 Calculate general term of expansion32𝑥2 − 13𝑥6 We know that general term of expansion (a + b)n is Tr+1 = nCr (a)n–r.(a)n For general term of expansion 32𝑥2 − 13𝑥6 Putting n = 6 , a = 32𝑥2 , b = −13𝑥 Tr+1 = 6Cr 32𝑥26 − 𝑟−13𝑥𝑟 = 6Cr 326 − 𝑟 (𝑥2)(6 − 𝑟) −13× 1𝑥𝑟 = 6Cr 326 − 𝑟 (𝑥)2(6 − 𝑟) −13𝑟1𝑥𝑟 = 6Cr 326 − 𝑟 (𝑥)12 − 2𝑟 −13𝑟𝑥−𝑟 = 6Cr 326 − 𝑟 −13𝑟 (𝑥)12 − 2𝑟 𝑥−𝑟 = 6Cr 326 − 𝑟 −13𝑟 (𝑥)12 − 2𝑟 − 𝑟 = 6Cr 326 − 𝑟 −13𝑟 (𝑥)12 −3𝑟 Step 2 We need to find the term independent of x So, power of x is 0 𝑥12 − 3𝑟 = x0 Comparing powers 12 – 3r = 0 12 = 3r 123 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 326 − 4 −13(4) (𝑥)12 −3(4) T5 = 6C4 322 134 (𝑥)12 −12 = 6C4 3222134 (𝑥)0 = 6C4 1223234 (1) = 6C4 122134−2 = 6C4 122132 = 6!4!6−4! 1419 = 6!4!2! 1419 = 6(5)(4)!4!2! 1419 = 5 12 Hence, the term which is independent of x is T5 = 512

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.